To get tomorrow's date as a datetime in Python, the easiest way is to use the Python timedelta() function from the datetime module.
datetime.date.today() + datetime.timedelta(days=1)
should do the trick
timedelta
can handle adding days, seconds, microseconds, milliseconds, minutes, hours, or weeks.
>>> import datetime
>>> today = datetime.date.today()
>>> today
datetime.date(2009, 10, 1)
>>> today + datetime.timedelta(days=1)
datetime.date(2009, 10, 2)
>>> datetime.date(2009,10,31) + datetime.timedelta(hours=24)
datetime.date(2009, 11, 1)
As asked in a comment, leap days pose no problem:
>>> datetime.date(2004, 2, 28) + datetime.timedelta(days=1)
datetime.date(2004, 2, 29)
>>> datetime.date(2004, 2, 28) + datetime.timedelta(days=2)
datetime.date(2004, 3, 1)
>>> datetime.date(2005, 2, 28) + datetime.timedelta(days=1)
datetime.date(2005, 3, 1)
No handling of leap seconds tho:
>>> from datetime import datetime, timedelta
>>> dt = datetime(2008,12,31,23,59,59)
>>> str(dt)
'2008-12-31 23:59:59'
>>> # leap second was added at the end of 2008,
>>> # adding one second should create a datetime
>>> # of '2008-12-31 23:59:60'
>>> str(dt+timedelta(0,1))
'2009-01-01 00:00:00'
>>> str(dt+timedelta(0,2))
'2009-01-01 00:00:01'
darn.
EDIT - @Mark: The docs say "yes", but the code says "not so much":
>>> time.strptime("2008-12-31 23:59:60","%Y-%m-%d %H:%M:%S")
(2008, 12, 31, 23, 59, 60, 2, 366, -1)
>>> time.mktime(time.strptime("2008-12-31 23:59:60","%Y-%m-%d %H:%M:%S"))
1230789600.0
>>> time.gmtime(time.mktime(time.strptime("2008-12-31 23:59:60","%Y-%m-%d %H:%M:%S")))
(2009, 1, 1, 6, 0, 0, 3, 1, 0)
>>> time.localtime(time.mktime(time.strptime("2008-12-31 23:59:60","%Y-%m-%d %H:%M:%S")))
(2009, 1, 1, 0, 0, 0, 3, 1, 0)
I would think that gmtime or localtime would take the value returned by mktime and given me back the original tuple, with 60 as the number of seconds. And this test shows that these leap seconds can just fade away...
>>> a = time.mktime(time.strptime("2008-12-31 23:59:60","%Y-%m-%d %H:%M:%S"))
>>> b = time.mktime(time.strptime("2009-01-01 00:00:00","%Y-%m-%d %H:%M:%S"))
>>> a,b
(1230789600.0, 1230789600.0)
>>> b-a
0.0
Even the basic time
module can handle this:
import time
time.localtime(time.time() + 24*3600)
For people who are dealing with servers Time Stamp
To get yesterday Time Stamp:
yesterdaytimestamp = datetime.datetime.today() + datetime.timedelta(days=-1)
To get Today Time Stamp:
currenttimestamp = datetime.datetime.now().timestamp()
To get Tomorrow Time Stamp:
tomorrowtimestamp = datetime.datetime.today() + datetime.timedelta(days=1)
To print:
print('\n Yesterday TimeStamp is : ', yesterdaytimestamp.timestamp(),
'\n Today TimeStamp is :', currenttimestamp,
'\n Tomorrow TimeStamp is: ', tomorrowtimestamp.timestamp())
The output:
Yesterday TimeStamp is : 1632842904.110993
Today TimeStamp is : 1632929304.111022
Tomorrow TimeStamp is : 1633015704.11103
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